Question: Solve for $t$, $ -\dfrac{t}{t^3} = -\dfrac{2}{3t^3} + \dfrac{7}{t^3} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $t^3$ $3t^3$ and $t^3$ The common denominator is $3t^3$ To get $3t^3$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{t}{t^3} \times \dfrac{3}{3} = -\dfrac{3t}{3t^3} $ The denominator of the second term is already $3t^3$ , so we don't need to change it. To get $3t^3$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{7}{t^3} \times \dfrac{3}{3} = \dfrac{21}{3t^3} $ This give us: $ -\dfrac{3t}{3t^3} = -\dfrac{2}{3t^3} + \dfrac{21}{3t^3} $ If we multiply both sides of the equation by $3t^3$ , we get: $ -3t = -2 + 21$ $ -3t = 19$ $ -3t = 19 $ $ t = -\dfrac{19}{3}$